Finding equation whose roots are square of the original but intially some of them were negative .

Question

Consider a cubic given as $x^3 + ax^2 + bx + c = 0$ which has roots $\alpha,-\beta,-\gamma$ , we need to find cubic equation whose roots are $\alpha^2 ,\beta^2 , \gamma^2$ is it possible to get it by transformation of roots ? By letting √y = x and substituting it ? If its not possible does vietas work? As such i tried both but was not able to fully get the equation.

Answer 1

You know that your cubic is $f_1:=(x-\alpha)(x+\beta)(x+\gamma)$, you want to compute the cubic $f_2:=(x-\alpha^2)(x-\beta^2)(x-\gamma^2)$. By the description I gave above of your cubic you know $a,b,c$ in terms of $\alpha,\beta,\gamma$, so you are asking if it is possible to describe the coefficients of the second equation in terms of these coefficients.

I suggest you using CoCoa or a program like this: Work in $\mathbb{Z}[a,b,c,\alpha,\beta,\gamma,a',b',c']$ with the term order of elimination (lex term order for example) s.t. a',b',c' have the lowest weight and the give the highest weight to the coefficients $a,b,c$.

Now set the ideal $I$ given by the relations between coefficients $a,b,c,a',b',c'$ and the roots $\alpha,\beta,\gamma$ and the squares. Compute the representative element for $a',b',c'$ in this setting.

If there is some relation between $a,b,c$ and $a',b',c'$ Cocoa will find it for you ;)

Otherwise I see only the pencil and paper solution!

Answer 2

Let mark $\beta=-\beta$ and $\gamma=-\gamma$. This does not influence equation that we need to obtain. Using Vieta's theorem: $$\alpha+\beta+\gamma=-a, \alpha\beta+\alpha\gamma+\beta\gamma=b, \alpha\beta\gamma=-c$$

Then we need to express $\alpha^2+\beta^2+\gamma^2=A$, $\alpha^2\beta^2+\alpha^2\gamma^2+\beta^2\gamma^2=B$ and $\alpha^2\beta^2\gamma^2=C$ in terms of $a$, $b$, $c$.

One can easily show that $$a^2=A+2b, c^2=C, b^2=B+2\cdot(-a)\cdot(-c)=B+2ac$$ $$A=a^2-2b, C=c^2, B=b^2-2ac$$

The resulting equation is $$x^3-(a^2-2b)x^2+(b^2-2ac)x-c^2=0$$